Question #312ef

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Answer 1 · 2016-09-11T17:45:25

$(10e)^x/(ln(10)+1)+e^(x+1)+C$

Expanding the integral, it becomes:

$=inte^x10^xdx+inte^(x+1)dx$

First working with the second integral, let $u=x+1$ so $du=dx$ :

$=inte^x10^x+inte^udu$

$=inte^x10^x+e^u$

$=inte^x10^x+e^(x+1)$

For the remaining integral, rewrite as follows:

$=int(10e)^xdx+e^(x+1)$

Notice that $10e$ is just a constant. Use the rule: $inta^xdx=a^x/ln(a)+C$

$=(10e)^x/ln(10e)+e^(x+1)$

Note that $ln(10e)=ln(10)+ln(e)=ln(10)+1$ :

$=(10e)^x/(ln(10)+1)+e^(x+1)+C$